How do you find a quadratic equation if you are only given the solution? Is it possible to have different quadratic equations with the same solution? Explain. Provide me with one or two solutions with which they must create a quadratic equation.

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YOUR QUADRATIC IS IN THE FORM OF AX^2 +BX+C

USE THE DISCRIMINANT TO FIND OUT THE NATURE OF THE ROOTS, NOW THE NATURE OF THE ROOTS WILL TELL YOU IF THE ANSWER IS RATIONAL, IRRATIONAL, IMAGINARY, OR IF THERE IS ONE OR TWO ANSWER.

SINCE YOU ARE ONLY WORRIED ABOUT THE NUMBER OF SOLUTIONS THEN THAT'S ALL WE'LL FOCUS ON.

THE DISCRIMINANT IS B^2 -4AC

IF YOUR ANSWER IS NEGATIVE, THEN THERE IS NO SOLUTION

IF YOUR ANSWER IS ZERO, THEN THERE'S ONE SOLUTION

IF YOUR ANSWER IS POSITIVE, THEN THERE'S TWO SOLUTIONS

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REGARDING How do you find a quadratic equation if you are only given the solution?

YOU NEED SUM OF THE ROOTS AND PRODUCT OF THE ROOTS

SUM OF THE ROOTS = -B/A

PRODUCT OF THE ROOTS = C/A

SO LETS SAY YOU HAVE A QUADRATIC W SOLUTIONS (SAME THING AS ROOTS) X=5 AND X= 7

SUM OF THE ROOTS WOULD BE 12 (JUST ADD THE ANSWERS

PRODUCT OF THE ROOTS WOULD BE 35 (MULTIPLY THE ANSWERS)

SO.... SUM OF THE ROOTS IS -B/A, BUT WE KNOW ITS 12.. WE CAN SAY -B/A = 12/1. I PUT THE 12 OVER 1 BC ITS A WHOLE NUMBER AND I JUST LIKE TO COMPARE FRACTIONS TO OTHER FRACTIONS, ITS A LITTLE EASIER.

THE PRODUCT OF THE ROOTS IS C/A OR 35, SO C/A = 35/1.

C/A = 35/1 -B/A = 12/1

WE CAN THEN SAY THAT A = 1, B = -12 (WATCH YOUR SIGNS) AND C = 35

THE QUADRATIC THAT GOES W THIS WOULD BE

Y=X^2 -12X+35.

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Is it possible to have different quadratic equations with the same solution?

YES, A SOLUTION IS JUST WHERE THE QUADRATIC CROSSES OVER THE X AXIS.

SO EASY EXAMPLE WOULD BE Y=X^2 AND Y=5X^2. THEY ARE DIFFERENT QUADRATICS AND THEY BOTH HAVE ONLY ONE SOLUTION, ZERO

Discriminant of a quadratic equation provides you with the indication of the number of solutions.

For the equation ax^{2}+bx+c=0, D=b^{2}-4ac is the discriminant.

1) D>0; there are two distinct solutions.

x_{1}={-b-√(b^{2}-4ac)}/(2a)=(-b-√D)/(2a)

x_{2}=(-b+√D)/(2a)

Example:

x^{2}-5x+6=0; D=5^{2}-4*6*1=25-24=1;

x_{1}=(5-√1)/2=2; x_{2}=(5+√1)/2=3;

2) D=0; There is one solution. Strictly speaking, two solutions become coincident, that is x_{1}=x_{2}=-b/(2a);

In this case the left-hand side of the equation is a complete square.

Example: x^{2}-4x+4=0; left hand side is simply (x-2)^{2}; x_{1}=x_{2}=2;

3) D<0; There are no real solutions. There are two imaginary solutions, though.

Example:

x^{2}-x+1=0; D=1^{2}-4*1*1=-3<0; No real solutions exist.

If two quadratic equations have identical solutions, then the equations are identical up to a factor.

Example:

x^{2}-5x+6=0 and 3x^{2}-15x+18=0; The second equation is just the first equation multiplied by 3.

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